BARUCH COLLEGE MATH 05 FALL 007 MANUAL FOR THE UNIFORM FINAL EXAMINATION Joseph Collison, Warren Gordon, Walter Wang, April Allen Materowski, Sarah Harne The final eamination for Math 05 will consist of two parts. Part I: Part II: This part will consist of 5 questions similar to the questions that appear in Part I of each sample eam. No calculator will be allowed on this part. This part will consist of 0 questions similar to the questions that appear in Part II of each sample eam. The graphing calculator is allowed on this part. There ma be a few new problem tpes on the eam that are not similar to the problems in the sample eams. If such problems appear, the will be similar to problems that ou have seen during the semester. GRADING: Each question will be worth points. Anone who gets or 5 questions correct will be assigned a grade of 00. No points are subtracted for wrong answers. CONTENTS OF THIS MANUAL: Page showing the sample questions that correspond to each section of the current tet. (When a section has been covered in class, the list indicates the problems that can be used in studing for the eam that includes that section during the semester.) TI-89 Facts for the Uniform Final Eamination. (This portion indicates the minimal calculator knowledge needed.) Sample Eam A (A A5) Sample Eam B (B B5) Sample Eam C (C C5) Sample Eam D (D D5) Sample Eam E (E E5) Answers to the problems.
Math 05 Tetbook Sections Corresponding to Sample Uniform Final Eam Questions Fall 007 Tetbook: Applied Calculus, Gordon, Wang, Materowski, Baruch College, CUNY Section Problems. (Etrem A, A9, B, B9, B5, C, C8, D, D, D9, D5, E, E, E A6, B6, C6, D. ( st Der. Test) A, B, C, D, E. (Concavit) A, A6, A0, B, B0, C, C9, D, D0, E, E, E0, E A7, A, B7, C7, C, D6, E. (Geom. Apps.) A, B, C, C0, D, E9.5 (Business Apps.) A5, A, B5, B, C7, C, D5, D, D, E5, E, E B8.6 (Linearization) A, B6, B, C6, C, D6, D5, D, E6, E5 A8, B9, C8, D7, D8, E5. (Inverses) A7, B7, C5. (Eponent. F cns) A8, B8, C8, D B. (Number e) A9, B9, C9 A9, C9, E6, E7. (Derivative e^) A0, A, B0, B, C0, C, D7, D6, E7 A0, A, B5, C0, D9, E8, E9.5 (Logarithms) A, B, C, E8, E6 A5, D0, E0.6 (Log. Props/Der) A, A, B, B, C, C, D8, D7, D, E9, E7, E, E5 A, B0, B, C, D, D, E.7 (Applications) B, B, C, D5, E 5. (Antiderivatives) (See 5. Problems) 5. (Apps. Antider.) A, B, C, D9, E0 5. (Substitution) A, B 5. (Appro. Are A5, B8 A, C, E 5.5 (Sigma and Are B5 5.6 (Definite Integral) A6, B6, C5, D0 5.7 (Subst. Def. Int.) A7, A5, B7, C, C6, C5, D, D8, E, E8, E 5.8 (Applications) A8, C7 A, C5, D
TI-89 FACTS FOR THE UNIFORM FINAL EXAMINATION The following information represents the minimal calculator usage students should be familiar with when the take the uniform final eamination. Instructors are epected to provide much more information in class. Limits can be evaluated with the TI-89 b using the limit function whose snta is limit(epression, variable, value). This function is found on the Home screen b using the F Calc ke. The calculus menu that appears when F is pressed is shown in Figure. The Figure limit function is selected b either pressing or using the down arrow cursor to highlight choice :limit and then pressing ENTER. Eample : Find lim 6 6 Solution: Select the limit function as indicated above. Complete the command line as follows and press ENTER: limit((-)/(6- ()),,6) Figure The answer is as shown in Figure. Eample : Find lim Δ 0 +Δ Δ Solution: In order to use the calculator for this problem it is best to think of Δ as a single letter such as t. That is, lim + t t 0 t Now enter on the command line the following: Figure limit((/(+t)-/)/t,t,0) The result is -/ as shown in Figure. Note that eample could have been done without a calculator if the limit requested was recognized as the definition of the derivative of f() = / for which f () = -/. The snta for the solve command and for finding the first or second derivative is: solve(equation, variable) d(epression, variable) d(epression, variable, ) (first derivative, f (), where f () = epression) (second derivative, f (), where f () = epression)
The solve command is obtained b pressing the F (Algebra menu) ke in the Home Screen as shown in Figure and then pressing ENTER to select choice :solve. The differentiation operator d is choice on the Home screen calculus menu shown in Figure. It can be more easil accessed b pressing the ellow nd ke and then 8 (the d appears above the 8 in ellow). The purple D appearing Figure above the comma ke cannot be used for this purpose. If an answer such as e or 5/6 appears when a decimal answer is desired, the command can be repeated with the green diamond ke pressed before pressing ENTER. Eample : The demand function for a product is p = 0 - ln(), where is the number of units of the product sold and p is the price in dollars. Find the value of for which the marginal revenue is 0. Solution: The revenue function is R() = p = (0 - ln()). The marginal revenue is the derivative, R (). This can be found b hand or b using the calculator command d(*(0-ln()),) obtained b pressing the kes nd 8 ( * ( 0 - nd ) ), ) ENTER Figure 5 as shown in Figure 5. The result is R () = 9 - ln(). To find the value of for which the marginal revenue is 0 the equation 9 - ln() = 0 must be solved. Press F (Algebr ENTER as shown in Figure. Then complete the command line as solve(9-ln()=0,) Pressing ENTER produces the result e 9. Since a decimal answer is desired, repeat the command b first pressing the green diamond ke before pressing Figure 6 ENTER. The final answer is 80.08 as shown in Figure 6 and can be rounded off to 80. 0 Eample : The function ht () = has eactl one inflection point. Find 0.t 5 0 + e + it. Solution: Recall that an inflection point occurs if h (t) = 0 and the concavit changes sign. So first the second derivative is found with the command d(0/(0+e^(-0.t+5)),t,) where e^( is obtained b pressing (e appears in green over the ke) Figure 7 Figure 7 is the result. Recall that to see the entire result the up cursor arrow button must be pressed and then the right cursor arrow button must be pressed until the rest of the result on the right can be seen as shown in Figure 8. Thus, t t 5.9(.057) ((.057).8) h"( t) = t 5 (0(.057) + e ) Move the cursor down to the command line again. The easiest wa to solve h (t) = 0 is as follows. First press F (Algebr and select choice so that Figure 8
the command line now onl shows solve( Press the up cursor arrow to highlight the epression for h (t) and press ENTER. The command line now shows the end of solve(above epression Complete the command b entering =0,t)ENTER Figure 9 The answer appears in Figure 9. Since onl one answer appears and it is known that an inflection point eists, there is no need to check to see if the concavit changes at t = 6.97. It onl remains to find the value of the original function at t = 6.97. To do so, enter 0/(0+e^(-0.*6.97+5)) on the command line to obtain h(6.97) =.5 as shown in Figure 0. Therefore, the inflection point is (6.97,.5). Figure 0 Eample 5: 0.0559 and.788 are the onl critical numbers for f () = e 5 ln(/). Determine if the critical point (0.0559, -.7) is a relative minimum, a relative maimum or neither. Solution : Recall that if f (0.0559) is positive the critical point is a relative minimum, if it is negative the point is a relative maimum, and if it is 0 the first derivative test must be used. The ke with the smbol on it means when or such that. Figure Now enter the following command: d(e^(5)*ln(/),,) =0.0559 Since f (0.0559) = -0.9 is negative as shown in Figure, the critical point is a relative maimum. Solution : A graph that clearl showed the point (0.0559, -.7) would reveal what was true for the critical point. Press F for the = screen. Enter the function Figure =e^(5)*ln(/) as shown in Figure If F (Zoom) and choice 6:ZoomStd are selected (so that the and values go from -0 to 0), the result is Figure. Notice that this reveals nothing about the critical point in question. So it is desirable to look at the point more closel. (If ou are epert at using ZoomIn, this approach can be used instead of the one shown. Just make sure the new center chosen has a value near -.7.) A good start might be to select values of between - and and values of between -6 and -. So press F (Window) and enter the following values. min=- ma= scl= min=-6 ma=- scl= as shown in Figure. Figure Figure
Then press F (Graph). Figure 5 is the result. Now press F (Trace) and then press the right cursor arrow a few times and observe values of and for each point on the graph shown. Clearl the high point is the critical point and therefore the critical point is a relative maimum. Figure 5 Students should also be familiar with using F5 (Math) in the graph window to determine the eact location of a relative maimum or relative minimum displaed on the graph. For the graph shown in Figure 5 press F5 (Math) to obtain the menu shown in Figure 6. Select choice :Maimum. In response the Lower Bound? request, just use the cursor movement arrows to move the blinking cursor to the left of the maimum and press ENTER. Then, for the Upper Bound? request, move the blinking cursor to the right of the maimum and press ENTER. Figure 7 is the result and indicates that (0.0559, -.709) is a relative maimum. Figure 6 Eample 6: Given f () = + - - 6 + 5 i) Find the critical numbers. ii) Find the critical points Figure 7 iii) All of the following are graphs of f() in different graphical windows. Which graph most accuratel portras the function (shows its relative etrema, asmptotes, etc.)? b) c) d) e) Solution: i) The critical numbers are the solutions to 0 = f () = + - 8-6. Enter the command solve(^+^-8-6=0,) into the calculator as shown in the screen on the right. The critical numbers are = -.5, -.06 or.7
ii) For the critical points the value of is needed for each of the critical numbers found. Enter the command ^+^-^-6+5 =-.5 into the calculator. Repeat this for the other critical numbers. (Pressing the right cursor arrow removes the command line highlight and positions the cursor at the end of the line. Then the backspace ke can be used to eliminate the -.5. Then enter the net critical number.) The screen shown indicates the critical points are (-.5, -50.6), (-.06, 0.5) and (.79, -86.9). iii) A graphing screen should be used that includes the critical points found. The values shown should include all values between - and. The values shown should include all values between -50 and. A reasonable window to choose would thus be min=-5 ma=5 scl= min=-5500 ma=500 scl=500 The graph shown on the right is the result. So the answer is d. Integration and the TI-89 Eample 7: Evaluate d + ( In the HOME screen, Integration is under F(calc)->: ( / ^+ ), ) Figure 8 shows the entr on the TI-89. (integrate Figure 8 (Note that ou must insert the constant of integration on our own.) Eample 8: Evaluate + d The procedure is the same as for indefinite integrals, but with etra arguments: F(calc)->: integration ( / ^+ ),,,) (The last arguments are the lower limit, followed b the upper limit.) Figure 9 shows the entr on the TI-89. Figure 9
PART - NO CALCULATORS PERMITTED SAMPLE EXAM A A. Find all of the points at which the graph of f () = + 5 has horizontal tangent lines. (,0) onl b) (,0) onl c) (,) and (-,0) d) (,0) and (,) e) (,) onl A. If the derivative of g() is g () = ( - ) ( - ) then the function is decreasing on (-, ) b) decreasing on (-, ) and increasing on (, ) c) decreasing on (-, ) and increasing on (, ) d) increasing on (-, ) and decreasing on (, ) e) decreasing on (-, ), increasing on (, ) and decreasing on (, ) A. The position equation for a moving bod is Find the time when the velocit is 0. t t+ 0 s = for t 0. 5 b) / c) 0 d) 5/6 e) / A. The sum of a number and twice another number is 0. Find the largest possible product. (Notice that our answer is not one or both of the numbers, it is the product.).5 b) 8 c) d).6 e) A5. Find the number of units to be produced in order to maimize revenue if the demand function is given b p = 600-0.0. 5,000 b) 0,000 c) 5,000 d) 0,000 e) 5,000 A6. On the interval 0 < <, for the graph shown,.0 f () > 0 and f () > 0 b) f () > 0 and f ()< 0 c) f () < 0 and f () > 0 d) f () < 0 and f () < 0 e) None of the above.5.0 0.5 0.0 0 0. 0. 0.6 0.8 A7. Suppose = f() and = g() are inverse functions with f() = 9 and f '() = 7 then g '(9) is 7 b) -7 c) /7 d) -/7 e) does not eist A8. A population of bacteria is decaing continuousl at a rate of 7% ever hour. If there are. million bacteria present at 9AM, which of the following gives the population of bacteria t hours after 9AM? Pt ( ) =.(.07) t b) Pt ( ) =.(0.9) t c) Pt ( ) =. t 0.07t d) Pt ( ) =.e e) Pt ( ) =.e 0.07t
A9. Which of the following best represents the graph of = e +? (All - and - scales are.) b) c) d) e) d A0. Evaluate ( e ) d e b) e c) d) e) none of these e e A. Evaluate log 7 b) / c) d) 5/ e) 9 d d A. Evaluate ( ln ) b) ln c) ln d) ln e) A. The marginal cost, in dollars is given b the equation C ' () = + 00. If the fied overhead cost is $000, then the total cost of producing 0 items is $000 b) $00 c) $500 d) $00 e) $500 A. Evaluate ( ) d ( ) 5 5 + c b) ( ) + c c) ( ) 0 5 + c d) ( ) 6 + c e) 5 5 ( ) ( ) + c 5 5 A5. The area of the region bounded b f() =, the lines = and = and the -ais is to be approimated b 0 rectangles using the right endpoint of each rectangle. The sum is k = 0 k 0 9 b) 0 k = k 0 0 c) 0 k= 0 k + 0 9 d) 0 k= k + 0 0 e) none of these 0
A6. Evaluate ( ) d. 0 8 b) c) d) 6 e) 0 A7. Evaluate d /5 b) -ln5 c) ln5 d) /5 e) A8. The consumer surplus for the demand and suppl functions + = and - + = is ½ b) / c) 5/ d) 7/ e) A9. Find the critical numbers for g() =. 0, ± b) 0, ± c) ± ()/ d) onl 0, - e) onl 0 A0. (0, 5) (i.e. = 0 and = 5) is the onl critical point of f() = 6 + + 5. You do not have to verif this. Determine what is true of (0, 5). (0, 5) is a relative maimum. b) (0, 5) is a relative minimum c) (0, 5) is a saddle point d) (0, 5) is not a relative etremum e) no conclusion is possible A. Find the maimum profit for the profit function P ( ) = + 6. 8 b) c) d) 0 e) A. Let the profit function for a particular item be P() = -0 + 60-00. Use differentials to approimate the increase in profit when production is increased from 5 to 6. 0 b) 0 c) 55 d) 60 e) 70 A. The equation of the tangent line to the curve defined b which = is f ( ) = e + + at the point at = + b) = 5 + c) = - 5 d) = - + 7 e) = 7 - ln+ ln A. = ln6 ln b) c) ln ln6 d) ln 7 e) ln A5. 0 8+ 9d= 65/ 76 b) 6/5 c) d) / e) ½
PART II - CALCULATORS PERMITTED A6. Find the absolute etrema of f( ) = + on the closed interval [0, ]. (0, ) and (, ) b) (, /5) and (-, -) c) (, /5) and (0, 0) d) (, ) and (-, -) e) (, ) and (0, 0) A7. Find the relative maimum value for the function f() = 5-5 = -.87 b) 5 =.87 c) -0 5 = -.6 d) 0 5 =.6 e) A8. The profit from manufacturing items is given b P() = -0.5 + 6-0. Find ΔP on [0, ]. $5.00 b) $5.50 c) $6.00 d) $70.00 e) $75.50 A9. A job offer consists of a $7,000 starting salar with a % increase each ear. To the nearest dollar, what will the salar be in 7 ears? $,6 (b) $5,50 (c) $6,95 (d) $8,9 (e) $8,66 A0. The maimum value of f() = e - is e - 0.68 b) e.78 c) 0.5 d) -e -.78 e) 0 A. Given ln( - ) = -, then to three decimal places, =.50 b).0 c).057 d).0 e) 5.057 A. If $5,000 per ear flows uniforml over an 8 ear period and earns % interest, compounded continuousl, then the present value, to the nearest dollar, is $5 b) $5,56 c) $5.08 d) $55,6 e) $7,05 A. For the function e h ( ) =, which of the following are true about the graph of = h( )? I. The graph has a vertical asmptote at = 0 II. The graph has a horizontal asmptote at = 0 III. The graph has a minimum point. None b) I and II onl c) I and III onl d) II and III onl e) I, II and III A. The graph of = is concave down for > 0 b) increasing for > 0 c) has = 0 as a horizontal asmptote d) concave up for > 0 e) has = 0 as a vertical asmptote
A5. The best fit logarithmic function passing through the points (, ), (, ) and (, 7), rounded to three decimal places is = - 5 b) =.9 + 0.969 ln c) = -.09 + 7. ln d) = 0. +.667 e) = -5.08 + 8.57 ln
PART - NO CALCULATORS PERMITTED SAMPLE EXAM B B. Find all of the critical numbers for f ( ) 9 =. 0 b) -, c) d) -, 0, e) - B. Find the absolute maimum of f () = - on the closed interval [-, ]. - b) - c) 6 d) e) 0 B. Solve f () = 0 for the function f() = 9 +. = b) = c) = -9 d) = e) = 0 B. Of all numbers (positive, negative and zero) whose difference is, find the two that have the minimum product. One of the numbers is: b) - c) - d) e) - B5. Find the Elasticit of demand when the price is for the demand function p =00 ½ b) c) - d) -5/8 e) - ½ B6. The linearization of f( ) = 5 + near 0 = is: = 0+ b) = + 5 c) = 7 d) 0 + e) 0 B7. f ( ) = + +. Find( f ) ( 6). b) / c) d) / e) -/6 B8. When writing the formula for an eponential function passing through the points (-, /9) and (, 6) using the form = ab, what would be the base, b? /9 b) 8 c) /8 d) e) / B9. Which of the following represents the amount in an account t ears after investing $000 at a 6% annual rate compounded monthl? d) 0.06 000 t 0.06 000 + t ( ) b) 000 + 0.06 t c) e) 000e 0.06( t ) 0.06 000 + t
B0. If e f( ) =, then f () = e b) e - c) e - ( - ) d) e ( - ) e) e - ( - ) B. Evaluate log 9 b) - c) 8 d) -8 e) 7/ B. Evaluate ( log0)( log 0) log ( )( ) log 0 + log 0 b) ( 0) c) + log d) log log5 e) none of these B. The marginal cost (in thousands of dollars) of producing hundred items is given b C ( ) =. If overhead is $500, find the cost function, C(). C() = 50 + C o b) C() = -500 c) C ( ) = + 500+ C d) C ( ) 500 e) = + C ( ) = 96 0 B. d = + ln + = c b) ln + + c c) ln + + c d) + + e) e ln c + + c B5. Evaluate 0 k = k 55 b) 65 c) 0 d) 50 e) 78 B6. Consider the function defined b = f(), whose graph is given in the accompaning figure. Suppose the area of the region indicated b A is 5, the area of B is and f ( d ) =, then the area of the Region C is b) c) d) e) none of these
B7. Evaluate ( + ) 0 d -80 b) -5 c) 7 d) -/6 e) -/ B8. Find the APPROXIMATE area between the graph of = and the -ais between = 0 and =. For our approimation, use rectangles of equal width and use the right-hand endpoint of each rectangle to determine the height of the rectangle. 6/ b) c) 0 d) e) 6 B9. Find an relative etrema of 7 f( ) = 5 = 0 b) = 7/5 c) = -7/ d) = 7 e) there is no relative etremum B0. Identif the -coordinate of an relative etrema of = ln( ). Relative min at = - b) Relative ma at = - c) Relative min at = /e d) Relative ma at = /e e) No relative etrema B. An apartment comple has 500 units available. At a monthl rent of $000 all the apartments are rented. For each $0 increase in rent, apartments become vacant. If is the number of $0 increases in rent, the monthl Revenue will be: (000 0 ) (500 + ) b) (000 + ) (500 0 ) c) (000 + 0 ) (500 ) d) (000 ) (500 + 0 ) e) (000 500 ) (0 ) B. If = +, find the value of the differential d corresponding to a change in of d = 0. when =. 0. b) 0.5 c) 0.5 d). e) 6.5 B. Find d/d if e + = 0 b) e e c) e d) e e e) e B. log 5 = 5 b) 5/ c) /5 d) -5 e) 5/ B5. Find the critical numbers for f() = + +. - onl b) 0 and - c) 0 onl d) + 6 e) There are no critical numbers.
PART II - CALCULATORS PERMITTED B6. The absolute maimum value of the function f() = 6 + 5 on the closed interval [0, ] is -7 b) -7 c) d) - e) 5 B7. What is the relative maimum of the function = f( ) = +? (-, ) b) (, 0) c) (0, 0) d) (, ) e) (,.6) B8. Given the total cost function C ( ) = 0.0 + + 00, find the value of for which the average cost is a minimum. ( -5. (b) 5. (c) 78.59 (d) -78.59 (e) 500 B9. For the function f = ( ) =, find Δ ( ) 6 ( ) ( ) ( ) Δ + Δ + Δ + Δ b) c) d d) 6 ( ) ( ) ( ) Δ + Δ + Δ e) 6 ( Δ ) + ( Δ ) + ( Δ ) 0 B0. Find the inflection point for gt () = 5 + t + (, 8) b) (.678, ) c) (0,.908) d) (.556,.890) e) (0.78, 7.9808) B. The demand function for a product is p = 00-0. ln() where is the number of units of the product sold and p is the price in dollars. Find the value of for which the marginal revenue is one dollar per unit. 97.5 b) 0.00007 c) 88.9 d) 90.5 onl e) 0.0 and 90.5 B. Find the best fit eponential function to three decimal places for the data given in the table below 5 7 0 5 0 0.0 = 6.88 b).79(.0) = c) = + + 0.07.67.00 d) =.00+.900 e) =.85(.099) B When a certain radioactive element decas, the amount, in milligrams, that remains after t ears can be 0.00t approimated b the function At () = ke, where k is a constant. Approimatel how man ears would it take for an initial amount of 800 milligrams of this element to deca to 00 milligrams? 7 b) 7 c) 69 d) 86 e)77
B If e k =, find k to decimal places 0.000 b).59 c). d) no solution e).68 B5 Find the equation of the straight line that is tangent to f ( ) = e at = = 5.7 8.55 b) =.78 c) = 5.7.78 d) =.78 e) = 5.7
PART - NO CALCULATORS PERMITTED SAMPLE EXAM C C. The graph of the function f( ) = + + has as its vertical asmptote(s) = - onl b) = onl c) = and = - d) No vertical asmptotes e) = onl C. Find the open interval(s) on which f( ) = + is decreasing. (-, -) and (, ) b) (0, ) c) (-, ) d) (-, 0) e) (-, ) C. Given = -, find the critical points. Then determine whether the are relative etrema (the second derivative test is easiest). Sketch the graph and pick the sketch that resembles our sketch the best. C and C5. The height in feet above the ground of a ball thrown upwards from the top of a building is given b s = -6t + 60t + 00, where t is the time in seconds C. What is the maimum height (in feet) of the ball? 5 b) 600 c) 00 d) 00 e) 60 C5. What is v ()? ft/sec b) sec c) -86 ft/sec d) ft/sec e) sec C6. For the function f( ) = = +, find d. Δ + Δ + Δ b) 6 + c) (6 + ) d d) 6 ( ) ( ) ( ) ( ) Δ e) ( Δ ) C7. For a production level of units of a commodit, the cost function in dollars is C = 00 + 00. The demand equation is p = 00-0.05. What price p will maimize the profit? $00 b) $50 c) $900 d) $500 e) $6000 C8. An eponential graph containing the points (, 5) and (, ) has the equation 5 ( ) b) ( ) c) 5 5 / d) 5 5 / e) none of these
C9. If e = and e = then e + 5 = 6/5 b) 0 c) 86 d) 66 e) d d 0 ( ) C0. ( e ) = 0e 9 b) e 0 c) 0e 0 d) ( ) 0 0 e e) none of these C. Evaluate ln 7 e 5 57 7/5 b) e / c) -5/7 d) e 75 / e) none of these C. Evaluate log z 5 ( log )( log ) 5 ( log )( logz ) b) log + log log + 5logz c) log + log log 5logz d) log + log log + 5logz e) log + log log 5logz t C. The acceleration of an object is given b the equation at ( ) = 5e + t. Determine its velocit function v(t) if v(0) = 6. t t t vt () = t+ 6 b) vt () = 5e + + t+ c) vt () = 5e t + t t+ d) () 5 t vt = e + t t+ 6 e) vt () = 5e t + t t+ C. Find the average value of f ( ) = on the interval [0, ] b) c) d) 8 e) C5. The area of the region bounded b f() =, g() = - and the -ais from = ½ to = is given b ( ( ) ) d b) ( / d c) / ) d + ( ) d / d) ( ) d + d / e) none of these
C6. Evaluate ln 0 e d ( 9 e ) b) ½ c) 8 d) e) C7. The producer surplus for the demand and suppl functions + = and - + = is ½ b) / c) 5/ d) 7/ e) C8. The critical numbers for = - are = 0 onl b) = -, 0, c) = -, onl d) = - onl e) = -, onl C9. If f() = 8 0 + -, find the value of f () (the second derivative) at =. 9 b) 6 c) 9 d) 76 e) 8 C0. A farmer wishes to construct adjacent enclosures alongside a river as shown. Each enclosure is feet wide and feet long. No fence is required along the river, so each enclosure is fenced along sides. The total enclosure area of all enclosures combined is to be 900 square feet. What is the least amount of fence required? 0 feet b) 0 feet c)0 feet d) 70 feet e) 60 feet C. Find the elasticit of demand when the price is for the demand function 5 p+ = 00 -/ b) -5/ c) - d) / e) C. For the function f( ) = = +, find Δ 6( Δ ) + ( Δ ) + ( Δ ) b) 6 + c) (6 + ) d d) ( Δ ) e) ( Δ ) C. Find the derivative of f() = e -. -e b) e - c) -e - d) e - e) e - C. If = e k then k = ln b) ln c) ln - ln d) e e) e C5. Use our knowledge of the definite integral (and our knowledge of the graph of the function = 6 ) in order to evaluate 8 6 d. 0-8 b) 0/ c) 6π d) π e) 6π
PART II - CALCULATORS PERMITTED C6. One critical number of the function = + is.6 b) -0.8 c) 0.75 d) e) -.7 C7. Find the relative maimum of f ( ) = + on the closed interval [0, 6]. 6 b) c) 6.00 d) - e) 0 C8. For the function = ( ) =, find ε f d) ( ) 6 ( ) ( ) ( ) Δ + Δ + Δ + Δ b) 6 ( ) ( ) ( ) Δ + Δ + Δ e) c) d 6 ( Δ ) + ( Δ ) + ( Δ ) C9. The half-life of a radioactive substance is 500 ears. How much of a 8 lbs. sample of this substance will remain after 9000 ears? lbs b) lbs c) lbs d) 5 lbs e) 6 lbs 0. C0. The function f( ) = 8e has eactl two points of inflection. The value of for one of these points of inflection is 8 b) -6 c) -0.8 d) 0 e) 5.6 C. For which value of does the graph of the function f() = e - ln have a horizontal tangent? Round our answer to the nearest hundredth. b).76 c).8 d).9 e) Never C. Given ln( - ) = -, then to three decimal places, =.50 b).0 c).057 d).0 e) 5.057 C. How much should be deposited into an account toda if it is to accumulate to $5000 in 7 ears and it earns.8% annual interest compounded quarterl? $7.50 b) $96.60 c) $5.00 d) $957.00 e) $989.80 6 C. Given g ( ) = +, find g ( ) -8 b) / 0.05 c) 0 d) / 6 0.0565 e) /0 0.0695 C5 If $0,000 per ear flows uniforml over a 0 ear period and earns.5% interest, compounded continuousl, then the present value, to the nearest dollar, is $5,85 b) $9,7 c) $75,8 d) $77,8 e) $87,9
SAMPLE EXAM D PART - NO CALCULATORS PERMITTED D. For what value(s) of is the derivative of = f( ) = equal to zero or undefined? = 0 onl b) = onl c) = - onl d) = -, onl e) = -, 0, D. The absolute maimum of the function f () = - 9 on the interval [-, ] is 0 b) c) - d) 9 e) -9 D. If R = - + - is the revenue function, where is the amount spent on advertising, then the point of diminishing returns (point of inflection) occurs when = b) = - c) = 6 d) = - e) Not enough information is given to decide D. Select the correct mathematical formulation of the following problem. A rectangular area must be enclosed as shown. The sides labeled cost $0 per foot. The sides labeled cost $0 per foot. If at most $00 can be spent, what should and be to produce the largest area? Maimize if 0 + 0 = 00 b) Maimize if 0 + 0 = 00 c) Maimize if + = 00 d) Maimize + if = 00 e) Maimize + if 00 = 00 D5. Find the values of for which the profit P() = - + 5 - is a maimum on [0, 5]. = 0 b) = c) = 5 d) = and 5 e) = 0 and 5 D6. For the function = ( ) = +, find ε f Δ + Δ + Δ b) 6 + c) (6 + ) d d) 6 ( ) ( ) ( ) ( ) Δ e) ( Δ ) D7. Find the derivative of f() = e e b) e + e c) + e d) + e e) e D8. Given log( + ) + log( - ) = log, then = - b) 5 c) d) - 5 e) - and 5
d D9. Solve: = d d) C 5 = 5 + b) = +C c) 5 5 ( ) = 5 + C e) = 5 5 = + C D0. Evaluate ( + 5) d 5 b) 08 c) 0 d) 8 e) 6 D. Find the area under the curve of = 9 on the interval [0, 9]. 8 b) -8 c) d) 9 e) -9 D. What is the absolute maimum value of f () = 75 on the closed interval [0, ]? 0 b) -50 c) 50 d) 6 e) 75 D. Solve the equation = 7 b) there is no solution c) - d) - e) - D. If the demand equation is given b p = 9, then the marginal revenue is 0 when = 9 b) = c) = 6 d) = e) = 5 D5. The linearization of f ( ) = 5 7 + near 0 = is: = 5 b) = 5+ c) = + 6 d) = 5 + e) = + 65 D6. Find the equation of the line tangent to the graph of = e - at the point where =. = e + b) = - c) = - d) = - 6 e) = 5-8 ln0 ln D7. = ln5 0 b) 5/ c) d) e) /5
D8. Which of the following will give a result of 7? d d 7 (t ) dt b) ( t t 7) dt c) 0 d dt ( 7) t t dt d) ( 7) d t t d d t e) d ( t t 7) d D9. Determine the -coordinate of the point(s), if an, at which the graph of has a horizontal tangent. f( ) = + + 0 9 = -, 9 b) None c) =.778,.60 d) = 0,, 9 e) = D0. The position of an object at an time t is given b st = t + t+. Find the velocit when t = () 8 0 0 8 b) - c) -6 d) e) 5/ D. Find the elasticit of demand when the price is 0 for the demand function 5 + 6 p = 00 - / b) - / c) / d) - 6/5 e) / D. For the function f( ) = =, find d ( ) 6 ( ) ( ) ( ) Δ + Δ + Δ + Δ b) c) d d) 6 ( ) ( ) ( ) Δ + Δ + Δ e) 6 ( Δ ) + ( Δ ) + ( Δ ) D. Find the derivative of f () = -6 ln 6/ b) -6/ c) -6/(ln ) d) 6/ e) -6/ D and D5. For the function f() = 9 + 5 D. Find the point(s) of inflection (, ). (, 7) onl b) (5, -5) onl c) (, 7) and (5, -5) d) (, -9) e) (, -) D5. Find the absolute etrema on the closed interval [0, ]. minimum is -9 and maimum is 0 b) minimum is -9 and maimum is 7 c) minimum is -5 and maimum is 7 d) minimum is -5 and maimum is 0 e) minimum is 0 and maimum is 7
PART II - CALCULATORS PERMITTED D6. Given f( ) = + 5+ 7 find f (). 8 b) 0 c) 59 d) 8/59 e) D7 and D8. The weekl profit that results from selling units of a commodit weekl is given b P = 50-0.00-5000 dollars. D7. Find the actual weekl change in profit that results from increasing production from 5000 units weekl to 505 units weekl. $0.00 b) $00.00 c) $99. d) $98.00 e) $8. D8. Use the marginal profit function to estimate the change in profit that results from increasing production from 5000 units weekl to 505 units weekl. $0.00 b) $00.00 c) $99. d) $98.00 e) $8. D9. = -/ -0. is a critical number for f ( ) = e. The critical point (-0., -0.97) is a relative minimum b) a relative maimum c) neither a relative maimum nor minimum d) a point of inflection e) a saddle point D0. The best fit logarithmic function passing through the points (, ), (, ) and (, 7), rounded to three decimal places is -5.08 + 8.57 ln b) 8.57-5.08 ln c).6 +.57 ln d).57 +.6 ln e) none of these D. Find the critical numbers for f() = e 0 ln(). 0.89 onl b) 0.08 and 0.89 c) 0.789 onl d) 0.789 and 0.89 e) There are no critical numbers D. Find the equation of the straight line that is tangent to f() = at the point (5, ). = + b) =.807-78.905 c) = 5.75 -.875 d) = - 8 e) =.5678-75.89 D. Given the demand and suppl functions three decimal places. p = and p = + determine the producer surplus to 0.85 b) 0.6 c).5 d).06 e).96
D Determine the absolute etrema of the function f ( ) = ( ) on the interval [ 0, ] 0 and b) 0 and 0.55 c) there is no absolute maimum d) 0 and 0. e) there are no absolute etrema D5 Given that the deca constant for Radium is 0.0008/ear, how long, to the nearest ear, does it take a sample to deca to 5% of its present mass? b) 50 c) 87 d) 785 e) 5
PART - NO CALCULATORS PERMITTED SAMPLE EXAM E E. The graph of = f() appears on the right. Estimate the points at which the absolute minimum and the absolute maimum occur on the interval [0, ], that is, 0. absolute minimum: (0, 0); absolute maimum: (, ) b) absolute minimum: (0, 0); absolute maimum: (, ) c) absolute minimum: (-, -); absolute maimum: (, ) d) absolute minimum: (, -5); absolute maimum: (-, 5) e) absolute minimum: (, ); absolute maimum: (, ) E. Find the open interval(s) on which f ( ) = + 7 is increasing. (-, -) or (, ) (i.e. < - or > ) b) (, ) onl (i.e > onl) c) (-, ) (i.e. - < < ) d) (-, -) onl (i.e. < - onl) e) (-, ) (i.e. all real ) E. The position of an object at an time t is given b t = st () = 8t + 0t+ 0. Find the acceleration when 8 b) - c) -6 d) e) 5/ E. The position of an object at an time t is given b is 0. st () = 8t + 0t+ 0. Find the time when the velocit 8 b) - c) -6 d) e) 5/ E5. Find the maimum profit for the profit function P() = - + 0 -. 0 b) 9/ c) (5 + 9)/ d) 7/ e) 67/8 E6. A square is measured and each side is found to be 5 inches with a possible error of at most.0 inches. Use DIFFERENTIALS to find the approimate error in computing the area of the square..6 b).06 c).006 d). e).0 E7. Find the derivative of f() = e -5 + 7. -5e -5 b) -5e -5 + 7 c) e -5 d) -5e -5 + 7 + e -5 + 7 e) e -5 + e -5 + 7
E8. Evaluate ln5 e 5 b) e c) 5 d) /5 e) 7 E9. Find the derivative of f( ) ln =. = 7 ( + + 7 7) b) + 7 + = 7 ( 7) c) 7 + = 7 ( 7) d) + 7 = 7 ( 7) e) 7 + = 7 ( 7) E0. Solve = 5 + C ' 5 =. b) = 5 + C c) 5 = + C d) 5 = + C e) 5 = + C E. Evaluate 8 t dt. b) c) 0 d) e) + E and E. For the function f() = - - E. Find all critical points (, ). (, ) b) (, -) c) (, 0) d) (-, -) e) (0, -) E. Find the absolute etrema on the closed interval [0, 5] (i.e 0 5). minimum is - and maimum is b) minimum is -7 and maimum is - c) minimum is -7 and maimum is d) minimum is -7 and maimum is 5 e) minimum is -0 and maimum is 0 E. The demand function and cost function for units of a product are Find the marginal profit when = 00. 60 p = and C = 0.65 + 00. $.5 per unit b) $.58 per unit c) $9.50 per unit d) $87.5 per unit e) $.65 per unit
E5. If f () = 00 and f '() = 6, estimate the value of f () for the function = f( ) 9 b) 88 c) 06 d) e) 8 E6. Given the function f ( ) = 5e, which of the following would be the graph of f (All - and -scales are.) ( )? ` b) c) d) e) E7. 8 ln ln e = 8 b) e / c) 8 d) 8 e) 8 d E8. Evaluate ln tdt d ln b) c) t d) + e) ln ln
E9. Select the correct mathematical formulation of the following problem. A farmer wishes to construct adjacent fields alongside a river as shown. Each field is feet wide and feet long. No fence is required along the river, so each field is fenced along sides. The total area enclosed b all fields combined is to be 800 square feet. What is the least amount of fence required? ( Minimize if + 5 = 800 (b) Minimize if + 5 = 00 (c) Minimize + 5 if = 800 (d) Minimize + 5 if = 00 (e) Minimize ( + ) if = 800 E0. If + = find d d b) c) d) - e) - E. Find the number of units that will minimize the average cost function if the total cost function is C = + 00 6 b) 0 c) 0 d) 0 e) 80 E. Find the derivative of = ln(+ ) 7 6 7 ln( + ) b) ln( + ) c) ( + ) ln( + ) 6 7 d) + e) 7 + E. The position of an object at an time t is given b st ( ) = t 5t+ 0. Find the acceleration when t = -0 b) c) 8 d) e) ½ E. Find the average value of f ( ) = + on the interval [0, ] 7/6 b) / c) d) / e) 5/6 E5. When epressed as a single logarithm, log( ) log( + ) log + log z = log ( ) + z b) ( ) + log z c) ( ) z log + d) log( ) z + e) none of these
PART II CALCULATORS PERMITTED E6. $00 is invested for 0 ears at an interest rate of the nearest dollar, the total amount that accumulates. $50 b) $5 c) $7 d) $08 e) $80 7 % compounded MONTHLY. Determine, to E7. Evaluate lim 0 ( e ) b) 0 c) d) undefined e) e 6 e E8 and E9. Given f( ) = 0, 000 E8. Find the critical numbers. 0.8 onl b) 0.8 and.9 c) d) 0.8 onl e) 0.8 and 5.9 E9. The graphs shown below displa different graphical windows. More than one ma actuall be graphs the function f() shown above. Pick the graph that most accuratel portras the function (shows all its relative etrema, asmptotes, etc.). b) c) c) d) e) E0. To three decimal places, log =.57 b).6 c).5 d).879 e) 5.7 E. If + 7 ln( + e ) f( ) = + then to two decimal places, f '() = -.67 b).69 c).86 d) -.6 e) none of these
00,000 E. If dollars are spent on advertising, then the revenue in dollars is given b R = 5 0.005 + where 0 5000. How much is spent on advertising at the point of diminishing returns (point of inflection)? $0,576 b) $00,000 c) $50,086.60 d) $000 e) There is no point of diminishing returns E. The area of the region bounded b f() =, the -ais and the line = and = 5 is approimated b n = rectangles. If the right endpoint is used for each of these rectangles, then the approimate area obtained is 0 b) c) d) 5 e) 6 E Given g ( ) = 5 + + + +, find g (0.5).9869 b) 97.5 c) -.8 d) 0 e).58 E5 Linearize f( ) = ( + + ).7 near = 0. + b).7 + c) + d).7 e)
ANSWERS A e B d C d D e E b A c B c C d D a E a A e B e C b D a E c A a B c C b D a E e A5 b B5 e C5 e D5 b E5 b A6 b B6 c C6 c D6 e E6 d A7 c B7 b C7 b D7 b E7 b A8 d B8 d C8 c D8 b E8 c A9 e B9 c C9 e D9 d E9 d A0 d B0 d C0 c D0 b E0 a A b B d C c D a E d A e B c C c D a E a A d B d C e D e E c A c B d C a D c E a A5 d B5 b C5 c D5 a E5 b A6 d B6 b C6 d D6 e E6 e A7 c B7 b C7 a D7 d E7 e A8 a B8 c C8 b D8 d E8 a A8 b B9 e C9 b D9 d E9 d A0 b B0 c C0 c D0 b E0 c A b B c C a D b E d A a B b C a D c E d A c B b C c D e E d A c B a C b D d E e A5 a B5 b C5 c D5 b E5 c A6 e B6 e C6 a D6 d E6 b A7 d B7 d C7 b D7 c E7 a A8 b B8 c C8 e D8 b E8 e A9 b B9 a C9 a D9 a E9 c A0 a B0 b C0 e D0 a E0 b A b B a C b D b E d A b B e C b D b E d A d B c C c D b E d A a B b C b D b E e A5 e B5 c C5 b D5 a E5 a